Question

Consider two independent tosses of a fair coin. Let $A$be the event that the first toss results in heads, let $B$be the event that the second toss results in heads, and let $C$be the event that in both tosses the coin lands on the same side. Show that the events $A$, $B$, and $C$ are pairwise independent—that is, $A$ and $B$ are independent, $A$ and $C$ are independent, and $B$ and C are independent—but not independent.

Short answer

All the all three events are not independent:

$P\left(ABC\right)=P\left(\right\{(H,H)\left\}\right)=\frac{1}{4}\ne \frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}=P\left(A\right)P\left(B\right)P\left(C\right)$

Step-by-step solution

Step 1:Given Information

Given that two independent tosses of a fair coin. Let $A$be the event that the first toss results in heads, let $B$be the event that the second toss results in heads, and let $C$ be the event that in both tosses the coin lands on the same side.

Step 2:Explanation

On the off chance that a fair coin is thrown twice freely there are 4 similarly reasonable events:

$(H,H)\text{- both tosses resulted in heads,}\Rightarrow P(H,H)=P\left(H\right)\cdot P\left(H\right)=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}$

$(H,T)\text{- first toss is heads, the second tails,}P(H,T)=\frac{1}{4}$

$(H,T)\text{- first toss is tails, the second heads}P(T,H)=\frac{1}{4}$

$(T,T)\text{- both tosses resulted in tails,}P(T,T)=\frac{1}{4}$

These events are all mutually exclusive.

Explanation of Defined Events

Defined events:

$P\left(A\right)=P\left(\right\{(H,H),(H,T)\left\}\right)=P\left(\right\{(H,H)\left\}\right)+P\left(\right\{(H,T)\left\}\right)=2\frac{1}{4}=\frac{1}{2}$

$P\left(B\right)=P\left(\right\{(H,H),(T,H)\left\}\right)=\frac{1}{2}$

$P\left(C\right)=P\left(\right\{(H,H),(T,T)\left\}\right)=\frac{1}{2}$

Step 4:Explanation of Characterization of Independence

Characterization of independence is that the probability of intersection is the result of the probabilities:

$P\left(AB\right)=P\left(\right\{(H,H)\left\}\right)=\frac{1}{4}=\frac{1}{2}\cdot \frac{1}{2}=P\left(A\right)P\left(B\right)\Rightarrow A\text{and}B\text{are independent}$

$P\left(BC\right)=P\left(\right\{(H,H)\left\}\right)=\frac{1}{4}=\frac{1}{2}\cdot \frac{1}{2}=P\left(B\right)P\left(C\right)\Rightarrow B\text{and}C\text{are independent}$

$P\left(AC\right)=P\left(\right\{(H,H)\left\}\right)=\frac{1}{4}=\frac{1}{2}\cdot \frac{1}{2}=P\left(A\right)P\left(C\right)\Rightarrow A\text{and}C\text{are independent}$

All events are independent in pairs.

Step 5:Final Answer

Thus all three events are not independent:

$P\left(ABC\right)=P\left(\right\{(H,H)\left\}\right)=\frac{1}{4}\ne \frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}=P\left(A\right)P\left(B\right)P\left(C\right)$